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Aug 30, 2017 at 9:07 comment added Matthew Daws @Mikael de la Salle Right, yes! That's a good observation. (So, as an example on $L^2(\mathbb R)$ this is absolutely fine, but I was hasty to claim it was immediately an example on $\ell^2(X)$ for some $X$).
Aug 28, 2017 at 13:35 comment added Mikael de la Salle Matt, the choice of a basis does not induce an isomorphism from $L^2(\mathbb R;E)$ to $\ell^2(E)$, i.e. is not a regular isomorphism between $L^2$ and $\ell^2$. However, your example is correct because $L^2$ embeds "regularly" in an ultraproduct of $\ell^2$.
Aug 9, 2017 at 11:12 vote accept duh
Aug 9, 2017 at 10:30 history answered Matthew Daws CC BY-SA 3.0